Oxidation numbers and redox reactions

Chemistry· 25 min read

1. Rules for Assigning Oxidation Numbers★★☆☆☆⏱ 8 min

📘 Definition

Oxidation Number

The hypothetical charge an atom would have if all bonds in the compound/ion were fully ionic (also called oxidation state)

Example:

The oxidation number of Na in NaCl is +1, Cl is -1

An element in its elemental state has an oxidation number of 0
A monatomic ion has an oxidation number equal to its charge
Fluorine always has oxidation number -1 in all compounds
Oxygen is usually -2, except peroxides (-1) and when bonded to F (positive)
Hydrogen is usually +1, except metal hydrides (-1)
Sum of oxidation numbers = 0 for neutral compounds, equals overall charge for ions

📐 Worked Example

Assign oxidation numbers to all elements in (a) $H_{2} SO_{4}$ (b) $Cr_{2} O_{7}^{2 -}$

1
For (a) $H_{2} SO_{4}$ : Let oxidation number of S = $x$ . Use rules: H = +1, O = -2. Sum to 0 for neutral compound:
2
$2(+1) + x + 4(-2) = 0 \\ 2 + x - 8 = 0 \\ x = +6$
3
Result: H = +1, S = +6, O = -2. For (b) $Cr_{2} O_{7}^{2 -}$ : Let Cr = $x$ . Sum equals overall charge -2:
4
$2x + 7(-2) = -2 \\ 2x - 14 = -2 \\ 2x = 12 \\ x = +6$
5
Result: Cr = +6, O = -2

Exam tip:

Always confirm if the species is a neutral compound or a charged ion when calculating the sum of oxidation numbers

2. Oxidation, Reduction and Redox Agents★★☆☆☆⏱ 7 min

Oxidation is an increase in oxidation number, corresponding to loss of electrons. Reduction is a decrease in oxidation number, corresponding to gain of electrons. An oxidising agent (oxidant) is reduced and oxidises another species, while a reducing agent (reductant) is oxidised and reduces another species.

📐 Worked Example

Identify which element is oxidised, which is reduced, and name the oxidising/reducing agents in the reaction: $Zn (s) + CuSO_{4} (a q) \to ZnSO_{4} (a q) + Cu (s)$

1
Assign oxidation numbers to all species:
2
Zn(s): 0, Cu $^{2 +}$ (CuSO $_{4}$ ): +2, Zn $^{2 +}$ (ZnSO $_{4}$ ): +2, Cu(s): 0
3
Compare changes: Zn goes from 0 → +2 (increase, so oxidised). Cu goes from +2 → 0 (decrease, so reduced).
4
Conclusion: Zn is the reducing agent (it is oxidised, reduces Cu $^{2 +}$ ), Cu $^{2 +}$ from CuSO $_{4}$ is the oxidising agent (it is reduced, oxidises Zn).

3. Writing Balanced Half-Equations★★★☆☆⏱ 8 min

🚫 No Calculator

Half-equations separate oxidation and reduction processes, showing electron transfer. They are used to build full balanced redox equations, and require balancing for both atoms and charge. The method differs slightly for acidic vs alkaline conditions.

📐 Worked Example

Write a balanced half-equation for the reduction of $MnO_{4}^{-}$ to $Mn^{2 +}$ in acidic solution

1
1. Write unbalanced equation for the species:
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$MnO_4^- \rightarrow Mn^{2+}$
3
1. Balance Mn: already balanced (1 Mn each side). Balance O by adding H $_{2}$ O to the O-deficient side:
4
$MnO_4^- \rightarrow Mn^{2+} + 4H_2O$
5
1. Balance H by adding H $^{+}$ (for acidic conditions) to the H-deficient side:
6
$8H^+ + MnO_4^- \rightarrow Mn^{2+} + 4H_2O$
7
1. Balance charge by adding electrons to the more positive side. Left charge = +8 -1 = +7, right charge = +2. Add 5e $^{-}$ to left:
8
$5e^- + 8H^+ + MnO_4^- \rightarrow Mn^{2+} + 4H_2O$
9
1. Check: all atoms balanced, total charge on both sides = +2, so half-equation is correct

4. Combining Half-Equations for Full Redox Equations★★★☆☆⏱ 7 min

🚫 No Calculator

To form a full balanced redox equation, the total number of electrons lost in oxidation must equal the total number gained in reduction. Adjust coefficients to equalise electrons, then add the half-equations and cancel common species.

📐 Worked Example

Combine the oxidation half-equation $Fe^{2 +} \to Fe^{3 +} + e^{-}$ with the MnO $_{4}^{-}$ reduction half-equation from the previous example to form a full balanced equation

1
1. Oxidation loses 1 e-, reduction gains 5 e-. Multiply oxidation half-equation by 5 to equalise electrons:
2
$5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-$
3
1. Add the two half-equations together:
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$5Fe^{2+} + 5e^- + 8H^+ + MnO_4^- \rightarrow 5Fe^{3+} + 5e^- + Mn^{2+} + 4H_2O$
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1. Cancel the 5 electrons on both sides, no other common species to cancel:
6
$5Fe^{2+} + 8H^+ + MnO_4^- \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$
7
1. Check: Atoms are balanced, total charge left = +10 +8 -1 = +17, right = +15 +2 = +17, so equation is balanced

✓ Quick check

Test your understanding:

What is the coefficient of $I_{2}$ when the reaction between $I^{-}$ and $Cr_{2} O_{7}^{2 -}$ (forming $I_{2}$ and $Cr^{3 +}$ in acidic solution) is fully balanced?
- A) 1
- B) 3
- C) 6
- D) 2
Reveal answer
B) 3 —
Oxidation half: $2 I^{-} \to I_{2} + 2 e^{-}$ , reduction half: $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$ . Multiply oxidation by 3 to get 6 electrons lost, so 3 $I_{2}$ is produced.

5. Common Pitfalls

Wrong move:

Assuming oxygen is always -2, even in peroxides

Why:

Common exception to the rule that is frequently tested in multiple choice questions

Correct move:

Check for peroxide (O-O) bonds or oxygen bonded to fluorine, and use -1 for oxygen in peroxides

Wrong move:

Using 0 as the sum of oxidation numbers for polyatomic ions

Why:

Students forget to account for the overall charge of the ion, leading to incorrect oxidation number calculations

Correct move:

Sum of oxidation numbers always equals the overall charge of the species, only 0 for neutral compounds

Wrong move:

Labelling the oxidised species as the oxidising agent

Why:

Common confusion between the process (oxidation/reduction) and the role of the species (agent)

Correct move:

Remember: Oxidising agents get reduced, reducing agents get oxidised

Wrong move:

Adding half-equations without equalising the number of electrons first

Why:

Skipping this step leads to unbalanced charge in the final full equation

Correct move:

Always cross-multiply half-equations so total electrons lost = total electrons gained before adding

Wrong move:

Leaving H+ in the final equation for reactions in alkaline solution

Why:

H+ does not exist in high concentration in alkaline conditions, so the equation is incorrect

Correct move:

Neutralise all H+ by adding equal OH- to both sides, then combine into water and simplify

6. Quick Reference Cheatsheet

Rule / Term	Key Value / Meaning
Elemental state	Oxidation number = 0
Monatomic ion	ON = ion charge
F in all compounds	ON = -1
O (not peroxide/F)	ON = -2
O in peroxides	ON = -1
H (not metal hydrides)	ON = +1
H in metal hydrides	ON = -1
Sum ON (neutral)	= 0
Sum ON (ion)	= overall ion charge
Oxidation	Increase in ON, lose e⁻
Reduction	Decrease in ON, gain e⁻
Oxidising agent	Gets reduced, gains e⁻
Reducing agent	Gets oxidised, loses e⁻

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 2
Assign oxidation numbers in MnO4-
2023 · 1
Identify reducing agent in reaction
2021 · 4
Balance full redox equation

What's Next

Mastering oxidation numbers and redox balancing is the foundation for all further electrochemistry topics in CIE A-Level Chemistry. You will apply these core skills to calculate standard cell potentials, predict reaction feasibility, and solve quantitative problems involving electrolytic cells and Faraday's laws of electrolysis. Redox concepts also appear regularly in organic chemistry, for example to identify oxidation and reduction processes in reactions of alcohols and carbonyl compounds. Building on this topic will prepare you for both multiple choice and structured extended response questions that make up a large proportion of exam marks.